Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/TRCSRSLASHEx6_Luc98

Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MARK₁(first₂(X1, X2)) -> MARK₁(X1)
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)
A__FIRST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

MARK₁(first₂(X1, X2)) -> MARK₁(X1)
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FROM₁(X) -> MARK₁(X)
A__FIRST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

MARK₁(first₂(X1, X2)) -> MARK₁(X1)
MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
MARK₁(from₁(X)) -> MARK₁(X)
MARK₁(from₁(X)) -> A__FROM₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
A__FIRST₂(s₁(X), cons₂(Y, Z)) -> MARK₁(Y)

The remaining pairs can at least by weakly be oriented.

A__FROM₁(X) -> MARK₁(X)

Used ordering: Combined order from the following AFS and order.
MARK₁(x1) = MARK₁(x1)
first₂(x1, x2) = first₂(x1, x2)
A__FIRST₂(x1, x2) = A__FIRST₁(x2)
mark₁(x1) = x1
s₁(x1) = s₁(x1)
from₁(x1) = from₁(x1)
A__FROM₁(x1) = A__FROM₁(x1)
cons₂(x1, x2) = cons₁(x1)
a__first₂(x1, x2) = a__first₂(x1, x2)
a__from₁(x1) = a__from₁(x1)
nil = nil
0 = 0

Lexicographic Path Order [19].
Precedence:

[MARK₁, A_{_FIRST₁}, A_{_{FROM_1]}}
[from₁, cons₁, a_{_{from_1]}} > [first₂, a_{_first₂}, nil]
[from₁, cons₁, a_{_{from_1]}} > s₁
0 > [first₂, a_{_first₂}, nil]

The following usable rules [14] were oriented:

mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
a__from₁(X) -> from₁(X)
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__first₂(X1, X2) -> first₂(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

A__FROM₁(X) -> MARK₁(X)

The TRS R consists of the following rules:

a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(mark₁(Y), first₂(X, Z))
a__from₁(X) -> cons₂(mark₁(X), from₁(s₁(X)))
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(mark₁(X))
mark₁(0) -> 0
mark₁(nil) -> nil
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.